\(x^2+2x+3\) là số chính phương

=>\(x^2+2x+3=k^2\left(k\in Z\right)\)

=>\(x^2+2x+1-k^2=-2\)

=>\(\left(x+1\right)^2-k^2=-2\)

=>(x+1-k)(x+1+k)=-2

=>(x+1-k;x+1+k)∈{(1;-2);(-2;1);(-1;2);(2;-1)}

TH1: x+1-k=1 và x+1+k=-2

=>x+1-k+x+1+k=1-2

=>2x+2=-1

=>2x=-3

=>\(x=-\frac32\) (loại)

TH2: x+1-k=-2 và x+1+k=1

=>x+1-k+x+1+k=1-2

=>2x+2=-1

=>2x=-3

=>\(x=-\frac32\) (loại)

TH3: x+1-k=-1 và x+1+k=2

=>x+1-k+x+1+k=-1+2

=>2x+2=1

=>2x=-1

=>\(x=-\frac12\) (loại)

TH4: x+1-k=2 và x+1+k=-1

=>x+1-k+x+1+k=-1+2

=>2x+2=1

=>2x=-1

=>\(x=-\frac12\) (loại)

Vậy: Không có số tự nhiên x nào thỏa mãn

hết cứu

\(G=-3x^2-5x+1\\ =-3\left(x^2+2.\dfrac{5}{6}x+\dfrac{25}{36}\right)+\dfrac{37}{12}\\ =\dfrac{37}{12}-3\left(x+\dfrac{5}{6}\right)^2\\ Vì:\left(x+\dfrac{5}{6}\right)^2\ge0\forall x\in R\\ Vậy:G_{max}=\dfrac{37}{12}.khi.x=-\dfrac{5}{6}\)

Lời giải:

$-F=5x^2+4x-3=5(x^2+\frac{4}{5}x+\frac{2^2}{5^2})-\frac{19}{5}$

$=5(x-\frac{2}{5})^2-\frac{19}{5}\geq \frac{-19}{5}$ với mọi $x$

$\Rightarrow F\leq \frac{19}{5}$

Vậy $F_{\max}=\frac{19}{5}$. Giá trị này đạt tại $x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}$

E = - 3\(x^2\) - \(x\) + 2

E = - 3.( \(x^2\) + 2.\(\dfrac{1}{6}\)\(x\) + \(\dfrac{1}{36}\)) + 2

E = -3.(\(x\) + \(\dfrac{1}{6}\))² + \(\dfrac{25}{12}\)

Vì (\(x+\dfrac{1}{6}\))² ≥ 0 ∀ \(x\) ⇒ -3.(\(x+\dfrac{1}{6}\))² ≤ 0 ⇒ -3(\(x+\dfrac{1}{6}\))² + \(\dfrac{25}{12}\) ≤ \(\dfrac{25}{12}\)

E_max = \(\dfrac{25}{12}\) ⇔ \(x\) = - \(\dfrac{1}{6}\) 

Ta có: \(F=-5x^2-4x+3\)

\(=-5\left(x^2+\frac45x-\frac35\right)\)

\(=-5\left(x^2+2\cdot x\cdot\frac25+\frac{4}{25}-\frac{4}{25}-\frac35\right)\)

\(=-5\left(x+\frac25\right)^2+5\cdot\left(\frac{4}{25}+\frac35\right)=-5\left(x+\frac25\right)^2+\frac{19}{5}\le\frac{19}{5}\forall x\)

Dấu '=' xảy ra khi \(x+\frac25=0\)

=>\(x=-\frac25\)

Ta có: \(E=-3x^2-x+2\)

\(=-3\left(x^2+\frac13x-\frac23\right)\)

\(=-3\left(x^2+2\cdot x\cdot\frac16+\frac{1}{36}-\frac{1}{36}-\frac23\right)\)

\(=-3\cdot\left\lbrack\left(x+\frac16\right)^2-\frac{25}{36}\right\rbrack=-3\left(x+\frac16\right)^2+\frac{25}{12}\le\frac{25}{12}\forall x\)

Dấu '=' xảy ra khi \(x+\frac16=0\)

=>\(x=-\frac16\)

Ta có: \(F=-5x^2-4x+3\)

\(=-5\left(x^2+\frac45x-\frac35\right)\)

\(=-5\left(x^2+2\cdot x\cdot\frac25+\frac{4}{25}-\frac{4}{25}-\frac35\right)\)

\(=-5\left(x+\frac25\right)^2+5\cdot\left(\frac{4}{25}+\frac35\right)=-5\left(x+\frac25\right)^2+\frac{19}{5}\le\frac{19}{5}\forall x\)

Dấu '=' xảy ra khi \(x+\frac25=0\)

=>\(x=-\frac25\)

Lời giải:

$-E=3x^2+x-2=3(x^2+\frac{x}{3})-2$

$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$

$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$

$\Rightarrow E\leq \frac{25}{12}$

Vậy $E_{\max}=\frac{25}{12}$. Giá trị này đạt được khi $x+\frac{1}{6}=0\Leftrightarrow x=\frac{-1}{6}$

\(C=-x^2-3x+4\)

\(=-\left(x^2+3x-4\right)\)

\(=-\left(x^2+3x+\frac94-\frac{25}{4}\right)\)

\(=-\left(x+\frac32\right)^2+\frac{25}{4}\le\frac{25}{4}\forall x\)

Dấu '=' xảy ra khi \(x+\frac32=0\)

=>\(x=-\frac32\)